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# The Quotient Rule - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

The quotient rule states that the derivative of a function h(x) where h(x) = f(x)/g(x) is h'(x) = (g(x)f'(x) - f(x)g'(x))/(g(x))^{2}.

For example, let h(x)=x^{2}/4x^{3}-7. Our functions f and g are f(x)=x^{2} and g(x)=4x^{3}-7. By using normal differentiating rules, we know that f'(x)=2x and g'(x)=12x^{2}. Plug these values into the formula: h'(x)=((4x^{3}-7)(2x) - (x^{2})(12x^{2}))/(12x^{2})^{2}. By expanding, we get h'(x)=(8x^{4}-14x - 12x^{4})/144x^{4} = (-4x^{4}-14x)/144x^{4}. Simplified, this becomes h'(x)=(-2x^{3}-7)/72x^{3}.

Let's do some problems with the quotient rule. So let's just recall that the quotient rule is how we differentiate a quotient of two functions; f and g. We'll call f the high function, and g the low function. Remember that the quotient rule is low d high minus high d low over the square of what's below.

So let's take a look at this problem. We want to differentiate y equals e to the x over x². So dy/dx, the derivative, is going to be low d high. So I need to put my fraction bar in. Low d high would be x² times the derivative of e to the x which is e to the x, minus high d low that's e to the x times the derivative of this guy which is 2x, over the square of what's below. That becomes x to the fourth.

This can be simplified a little bit. I can pull an e to the x out. I'm left with x² minus 2x. Now one of those x's will cancel with the denominator. So this x, this x, and that guy. So I'm left with an x minus 2 all over x³. So there is your answer.

Let's try another one. I have y equals x² plus 3x plus 2 over 5 minus x. Again, the derivative is low d high. So this is the low function. D high would be 2x plus 3 minus high d low. The high function is x² plus 3x plus 2. D low would be -1, the derivative of 5 minus x, -1 over the square of what's below. So 5 minus x².

I can simplify this. I don't want to, it looks terrible. 5 minus x, let's see we're going to get -2x². We're going to get 10x minus 3x. So plus 7x and then we're going to get +50. Then here, we'll get minus, minus plus x². We're going to get plus 3x. We're going to get plus 2 all over 5 minus x². I'll leave the denominator that way.

Just to combine like terms in the numerator; I'm going to have minus x². I'm going to have 10x, and I'm going to have plus 17 all that over (5 minus x)². That's the quotient rule.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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